THEOREM 1.39 Now we need to consider the e arrows. To do so, we set up an extra bit of notation. For any state R of M, we define E(R) to be the collection of states that can be reached from members of R by going only along e arrows, including the members of R themselves. Formally, for RC Q let Every nondeterministic finite automaton has an equivalent deterministic finite automaton. PROOF Let N = (Q, E, 6, qo, F) be the NFA recognizing some language A. Ne construct a DFA M = (Q', £, 8', qo', F") recognizing A. Before doing the full onstruction, let's first consider the easier case wherein N has no e arrows. Later E(R) = {q\ q can be reached from R by traveling along 0 or more e arrows}. Then we modify the transition function of M to place additional fingers on all states that can be reached by going along e arrows after every step. Replacing 6(r, a) by E(8(r, a)) achieves this effect. Thus ve take the e arrows into account. 1. Q' = P(Q). Every state of M is a set of states of N. Recall that P(Q) is the set of subsets of Q. 2. For RE Q' and a € E, let 8'(R, a) = {q € Q] q € 6(r, a) for some r € R}. If R is a state of M, it is also a set of states of N. When M reads a symbol a in state R, it shows where a takes each state in R. Because each state may go to a set of states, we take the union of all these sets. Another way to write this expression is 8'(R,a) = {q € Qlq € E(6(r, a)) for some r € R}. Additionally, we need to modify the start state of M to move the fingers ini- tially to all possible states that can be reached from the start state of N along the e arrows. Changing 4o' to be E({q}) achieves this effect. We have now completed the construction of the DFA M that simulates the NFA N. The construction of M obviously works correctly. At every step in the com- putation of M on an input, it clearly enters a state that corresponds to the subset of states that N could be in at that point. Thus our proof is complete. 8 (R,a) = U 6(r,a).* rER
THEOREM 1.39 Now we need to consider the e arrows. To do so, we set up an extra bit of notation. For any state R of M, we define E(R) to be the collection of states that can be reached from members of R by going only along e arrows, including the members of R themselves. Formally, for RC Q let Every nondeterministic finite automaton has an equivalent deterministic finite automaton. PROOF Let N = (Q, E, 6, qo, F) be the NFA recognizing some language A. Ne construct a DFA M = (Q', £, 8', qo', F") recognizing A. Before doing the full onstruction, let's first consider the easier case wherein N has no e arrows. Later E(R) = {q\ q can be reached from R by traveling along 0 or more e arrows}. Then we modify the transition function of M to place additional fingers on all states that can be reached by going along e arrows after every step. Replacing 6(r, a) by E(8(r, a)) achieves this effect. Thus ve take the e arrows into account. 1. Q' = P(Q). Every state of M is a set of states of N. Recall that P(Q) is the set of subsets of Q. 2. For RE Q' and a € E, let 8'(R, a) = {q € Q] q € 6(r, a) for some r € R}. If R is a state of M, it is also a set of states of N. When M reads a symbol a in state R, it shows where a takes each state in R. Because each state may go to a set of states, we take the union of all these sets. Another way to write this expression is 8'(R,a) = {q € Qlq € E(6(r, a)) for some r € R}. Additionally, we need to modify the start state of M to move the fingers ini- tially to all possible states that can be reached from the start state of N along the e arrows. Changing 4o' to be E({q}) achieves this effect. We have now completed the construction of the DFA M that simulates the NFA N. The construction of M obviously works correctly. At every step in the com- putation of M on an input, it clearly enters a state that corresponds to the subset of states that N could be in at that point. Thus our proof is complete. 8 (R,a) = U 6(r,a).* rER
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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